Homework #4 1. In general, the internal energy U depends on both temperature and volume, U=U(T,V). The volume dependence comes from the potential energy due to the interactions among the particles. For free particles, there are no interactions, thus, the ∂U internal energy U should be independent of the volume, i.e., = 0 . Verify this ∂ V T result for ideal gas pV = RT . ∂U ∂S From TdS = dU + pdV , or dU = TdS − pdV , there is = T − p . ∂ ∂ V T V T ∂S ∂p ∂U ∂P Using Maxwell relation = , we have = T − p . ∂ ∂ ∂ ∂ V T T V V T T V For ideal gas pV = RT or p = RT /V . ∂U ∂P R Thus = T − p = T − p = p − p = 0 . ∂ ∂ V T T V V ∂C ∂ 2 P ∂C ∂ 2V 2. Prove V = T and P = −T . ∂ ∂ 2 ∂ ∂ 2 V T T V P T T P For ideal gas, pV = RT . Show that CV is independent of the volume and CP is independent of the pressure. ∂ ∂ ∂ ∂ ∂ ∂ = S CV = S = S Since CV T , there is T T . ∂ ∂ ∂ ∂ ∂ ∂ T V V T V T V T T V T V ∂S ∂p Using Maxwell relation = , ∂ ∂ V T T V ∂C ∂ ∂S ∂ ∂p ∂ 2 p we have V = T = T = T . ∂ ∂ ∂ ∂ ∂ ∂ 2 V T T V T V T T V V T V ∂S Similarly, from C = T , there is P ∂ T p ∂C ∂ ∂ ∂ ∂ p S S = T = T . ∂p ∂p ∂T ∂T ∂p T p T T p ∂S ∂V Using Maxwell relation = − , ∂ ∂ p T T p ∂C ∂ ∂S ∂ ∂V ∂ 2V p = = − = − we have T T T 2 . ∂p ∂T ∂p ∂T ∂T ∂T T T p p p p RT RT For ideal gas, p = and V = . V p ∂p R ∂2 p ∂V R ∂ 2V Then = , = 0 and = , = 0 . ∂ ∂ 2 ∂ ∂ 2 T V V T V T p p T p 2 ∂ 2 ∂C ∂ p C p ∂ V Thus V = T = 0 and = −T = 0. ∂ ∂ 2 ∂ ∂ 2 V T T V p T T p 3. For free particles in two dimensions, what is density of states (DOS) in low speed limit (ε=p2/2m), and in high speed limit (ε=pc)? A2πpdp For a 2D system, D(ε )dε = , where A is the area of the system. h2 p 2 pdp (a) For ε = , dε = or pdp = mdε . Thus 2m m A2πpdp A2πmdε A2πm D(ε )dε = = or D(ε ) = . h 2 h 2 h 2 dε (b) For ε = pc , dε = cdp or dp = . Thus c A2πpdp A2πεdε A2πε D(ε )dε = = or D(ε ) = . h2 h2c 2 h2c2 4. If a kind of particles obeys the following dispersion ε=Dpα, what is the density of states (three dimensions)? V 4πp 2dp D(ε )dε = . Here V is the volume of the system. h3 1 α − α 1/ 1 α ε ε dε For ε = Dp , or p = and dp = . Thus D αD1/α 2 α− α − V 4πp 2dp V 4π ε α ε 1/ 1dε V 4π ε 3/ 1dε D(ε )dε = = = ⋅ . h3 h3 D αD1/α h3 αD3/α 4πVε 3/α −1 D(ε ) = . h3αD3/α 5. A system consists of N simple harmonic oscillators (one dimension). In the classical p 2 mω 2 x 2 limit, the energy of a simple harmonic oscillator is ε = + . Get the 2m 2 partition function, entropy, and the internal energy of this system. The single particle partition function is β 2 β ω 2 2 = dxdp −βε = dxdp − p − m x Z1 ∫ e ∫ exp h h 2m 2 +∞ +∞ 1 βp 2 βmω 2 x 2 = ∫ exp− dp ∫ exp− dx h −∞ 2m −∞ 2 +∞ +∞ 4 βp 2 βmω 2 x 2 = ∫ exp− dp ∫ exp− dx h 0 2m 0 2 β βmω 2 Let y = p and z = x . 2m 2 +∞ β 2 +∞ β ω 2 2 = 4 − p − m x Z1 ∫ exp dp ∫ exp dx h 0 2m 0 2 +∞ +∞ 4 2m 2 = ⋅ exp()− y 2 dy exp ()− z 2 dz β β ω 2 ∫ ∫ h m 0 0 π π π = 4 2m ⋅ 2 ⋅ ⋅ = 2 = 1 = kBT h β βmω 2 2 2 hβω βω ω The partition function is k T N Z = Z N = B . 1 ω k T The free energy is F = −k T log Z = −Nk T log B . B B ω ∂ log Z ∂ N The energy is U = − = − [−N log()βω ] = = Nk T . ∂β ∂β β B U − F 1 k T k T The entropy is S = = Nk T + Nk T log B = Nk 1 − log B . T T B B ω B ω .